Identifying the recurrence relation simply by iteration and a
good guess is often inadequate for even slightly complex relations. A
variety of techniques is available for finding explicit formulas for
special classes of recursively defined sequences. The method explained
here works for the Fibonacci and other similarly defined sequences.

Definition
A second-order linear homogeneous recurrence relation with
constant coefficients is a recurrence relation of the form ak
= A×ak
- 1
+ B×ak
- 2
for all integers k³
some fixed integer, where A and B are fixed real numbers with B ¹ 0.

"Second-order" refers to the fact that the expression of ak
contains the two previous terms ak -
1 and ak
- 2 with B ¹ 0, "linear" to the fact that ak-1
and ak-2
appear in separate terms and to the first power, "homogeneous" to the
fact that the total degree of each term is the same (no ak0or more commonly, no constant term), and "constant
coefficients" to the fact that A and B are fixed real numbers that do
not depend on k.

Example
State whether each of the following is a second-order linear
homogeneous recurrence relation with constant coefficients:

ak

= 3×ak
- 1 + 2×ak -
2

bk

= bk -
1 + bk -
2 + bk -
3

mk

= 1/2×mk
- 1 - 3/7×mk -
2

tk

= 2×tk
- 2

ak

= 3×ak
- 1 + k×ak -
2

ak

= 3×ak
- 1 + 2×ak -
2 + 1

Lemma
5.8.1
Let A and B be real numbers. A recurrence relation of the form

ak
= A×ak - 1
+ B×ak - 2

is satisfied by the
sequence 1, t, t2,
t3, ..., tn,
... where t is a nonzero real number, if, and only
if, t satisfies the equation

t2
- At - B = 0.

The equation t2
- At - B = 0 is called the characteristic
equation of the recurrence relation.

are the two solutions of
this form. Additional solutions can be found by taking linear
combinations of these two solutions. Say rn
=1, r, r
2, r 3, ..., r
n, ... and sn=
1, s, s 2,
s 3, ..., s
n, ... both satisfy the same second-order linear
homogeneous recurrence relation with constant coefficients, and C and D
are numbers then another sequence satisfying this recurrence relation
is defined by the formula

Now store these values for
C and D in your calculator. Let's test our results:

an
= C×rn
+ D×sn,
for all integers n³ 0

ak = 3×ak -
1 + 2×ak
- 2 for all integers
k ³2

a0 = C×r0
+ D×s0
= C + D = 1

a1 = C×r + D×s = 8

a2 = C×r2
+ D×s2
= 26

26 = 3× 8
+ 2× 1

a3 = C×r3
+ D×s3
= 94

94 = 3× 26
+ 2× 26

a4 = C×r4
+ D×s4
= 334

334 = 3×
94 + 2× 26

a25 = C×r25
+ D×s25
= 1.28325450305 x 1014

Example
A Formula for the Fibonacci Sequence
The Fibonacci sequence satisfies the recurrence relation

Fk
= Fk - 1
+ Fk - 2,
for all integers k³
2

with initial conditions

F0 = F1
= 1.

Find an explicit formula for this sequence.Solution: Certainly the Fibonacci
relation is a second-order linear homogeneous recurrence relation with
constant coefficients. We need to determine if its characteristic
equation has two distinct roots to satisfy the conditions of the
Distinct Roots Theorem. The characteristic equation is t2
- t - 1 = 0, which has solutions:

By the Distinct Roots Theorem, the Fibonacci sequence
satisfies the explicit formula

for
all integers n³ 0.

To find C and D we need
only solve the system of equations 1 = C + D and 1 = C×r + D×s, where r
and s are the roots of the characteristic equation.
Substitute C = 1 - D into the second equation to get